Quote:
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Originally Posted by sanddrag
Not enough torque. Need I say more? Actually, I shouldn't jump to such a conclusion. You never said what size wheels you were using. But assiming you wanted to go 112 mph, you would need a wheel approximately 2" in diameter on the 19670 rpm drill shaft. Now, at stall the wheel would only be able to exert approximately 7.9 lbs of force on the ground. As motor speed goes up, torque goes down. So, I reckon moving at 112 mph (near 19670 rpm), you will have very little torque, certainly not enough to counter the air resistence.
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Well, there is a really simple explanation to why you are almost definitley wrong in your reasoning. Unless the drill is much less efficient in (input/output
and output/weight), then a direct drive configuration is probably precisely what you are looking for. If a motor, batteries, and chasis can go at 111 mph, then a motor with an output/weight ratio anywhere near the ones used in the 111 mph car would, in at least some configuration, be able to travel at 111 mph also. Using your calculations, we see that to achieve this with just a drill motor and a wheel, we need to actually use
larger wheels and therefore
less force at the wheel because that is using the free speed which is no where near the peak power output (and therefore a very bad output/weight ratio). Using the formula for drag in a non-compressing fluid:
D=(1/2)*CD*A*
r*V2
we see that the drag force of such a car (assuming it is a sphere with radius 10cm)
= (1/2)*(0.5)*3.14*(0.1 m)2*(1.2 kg/m3)*(49.6m/s)^2 = 23.2 N or ~5lb.
The drill at max power is at ~10k rpm therefore requiring about 4" wheels and outputting about 7.9/(2*2) = ~2 lb. (divided by 4 because double size wheels and half stall torque). $@#$@#$@#$@#, maybe I should have made those calculations before starting to write this
. Seriously though, I think this result shows that is reasonable to have the drill motor going at 111 mph on direct drive, especially because of the highly overestimate drag (a sphere the size of a basketball?).